Use the mass of the W^- particle to estimate the range of the weak interaction responsible for the neutron beta decay. Strategy This calculation is much like the one we did in Example 14.1, when we discussed Yukawa’s meson. In that case we knew the nuclear force range, but in this case, the force

Question

Use the mass of the [latex]W ^{-}[/latex] particle to estimate the range of the weak interaction responsible for the neutron beta decay.

Strategy This calculation is much like the one we did in Example 14.1, when we discussed Yukawa’s meson. In that case we knew the nuclear force range, but in this case, the force range is unknown. We use Equation (14.4) to find the force range [latex]R_{ W }[/latex].

[latex]R=\frac{\hbar}{2 m c}=\frac{\hbar c}{2 m c^{2}}[/latex] (14.4)

Accepted Answer

The range [latex]R_{ W }[/latex] becomes

[latex]\begin{aligned}R_{ W } &=\frac{\hbar c}{2 m_{ W } c^{2}} \&=\frac{1.973 imes 10^{2} eV \cdot nm }{2\left(80.4 GeV / c^{2} ight)\left(c^{2} ight)}=1.2 imes 10^{-18} m\end{aligned}[/latex]

In this case, it may not be true that the [latex]W ^{-}[/latex] travels near the speed of light, because it is such a massive particle. The calculation here is therefore an upper limit. Note that in this case the violation of the conservation of energy is extreme, so the violation must occur over a very short period of time. We calculate [latex]\Delta t[/latex] from Equation (14.2) to be

[latex]\Delta t \approx \frac{\hbar}{2 \Delta E}=\frac{\hbar}{2 m_{\pi} c^{2}}[/latex] (14.2)

[latex]\begin{aligned}\Delta t &=\frac{\hbar}{2 m_{ W } c^{2}} \\Delta t &=\frac{1.055 imes 10^{-34} J \cdot s }{2(80.4 GeV )}\left(\frac{1 GeV }{1.6 imes 10^{-10} J } ight) \approx 4 imes 10^{-27} s\end{aligned}[/latex]

The lifetime of the neutron is much longer than the time it takes for the decay process itself.

Question 14.2: Use the mass of the W^- particle to estimate the range of th...

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